pku 1661 Help Jimmy

#include <iostream>#include <climits>#include <algorithm>using namespace std;struct Board{int left, right;int height;};Board board[1005];int Left[1005], Right[1005];int n;int N, X, Y, MAX;#define INTMAX INT_MAX/10bool cmp(const Board& lhs, const Board& rhs){if(lhs.height > rhs.height)return true;else if(lhs.height == rhs.height && lhs.left < rhs.left)return true;elsereturn false;}inline int min(const int& lhs, const int& rhs){return lhs < rhs ? lhs:rhs;}inline bool notBeCovered(int p, int i, int j){for(int k = i-1; k >= j+1; --k)if(board[k].left <= p && board[k].right >= p)return false;return true;}inline bool canGoToGroud(int p, int i){for(int k = i+1; k <= N; k++)if(board[k].left <= p && board[k].right >= p)return false;return true;}int main(){scanf("%d", &n);while(n--){scanf("%d%d%d%d", &N, &X, &Y, &MAX);for(int i = 1; i <= N; ++i)scanf("%d%d%d", &board[i].left, &board[i].right, &board[i].height);sort(board+1, board+N+1, cmp);memset(Left, 0x7777, sizeof(Left));memset(Right, 0x7777, sizeof(Right));board[0].left = board[0].right = X;board[0].height = Y;Left[0] = Right[0] = 0;for(int i = 1; i <= N; ++i){int min1 = INTMAX, min2 = INTMAX;for(int j = i-1; j >= 0; j--)if(board[j].height - board[i].height > 0&& board[j].height - board[i].height <= MAX){int temp1 = INTMAX;if(board[j].left <= board[i].right&&board[j].left >= board[i].left&& notBeCovered(board[j].left, i, j))temp1 = min(temp1, board[j].left-board[i].left+Left[j]);if(board[j].right <= board[i].right&&board[j].right >= board[i].left&& notBeCovered(board[j].right, i, j))temp1 = min(temp1, board[j].right-board[i].left+Right[j]);if(temp1 < min1)min1 = temp1;int temp2 = INTMAX;if(board[j].left <= board[i].right&&board[j].left >= board[i].left&& notBeCovered(board[j].left, i, j))temp2 = min(temp2, board[i].right-board[j].left+Left[j]);if(board[j].right <= board[i].right&&board[j].right >= board[i].left&& notBeCovered(board[j].right, i, j))temp2 = min(temp2, board[i].right-board[j].right+Right[j]);if(temp2 < min2)min2 = temp2;}Left[i] = min1, Right[i] = min2;}int Min = INTMAX;for (int i = 0; i <= N; ++i){if(board[i].height <= MAX){if(canGoToGroud(board[i].left, i) && Left[i] < Min)Min = Left[i];if(canGoToGroud(board[i].right, i) && Right[i] < Min)Min = Right[i];}}printf("%d/n", Min+Y);}return 0;}/*思路相对比较简单:设到达第i个板左端的最短时间Left[i], 到达第i个板右端的最短时间Right[i];所求即为每条可以到达地面的板(包括初始位置)的最小值 (其中要刨去超过MAX的).//此题需要注意的地方颇多:1. INTMAX 不可取INT_MAX, 否则INT_MAX再加上一个数可能溢出变成负数, 以后做题时也一定要注意!!!2. 注意Jimmy可以直接到地面的情形, 即:for (int i = 1; i <= N; ++i) => for (int i = 0; i <= N; ++i)3.要判断Jimmy初始位置能否到达该板:for(int i = 1; i <= N; ++i){int min1 = INTMAX, min2 = INTMAX;for(int j = i-1; j >= 0; j--)...}4. 判断每个板是否能否直接到地面时, 不要忘记高度限制, 我就在这WA了2次:for (int i = 0; i <= N; ++i){if(board[i].height <= MAX)....}*/